What is the derivative of f(x)=4x^{4/3}-2/3 x^{3/2}+x^2-4x+1 ?

The derivative of f(x)=4x^{4/3}-2/3 x^{3/2}+x^2-4x+1 is 2x+(16x^{1/3})/3-x^{1/2}-4

What is the first derivative of f(x)=4x^{4/3}-2/3 x^{3/2}+x^2-4x+1 ?

The first derivative of f(x)=4x^{4/3}-2/3 x^{3/2}+x^2-4x+1 is 2x+(16x^{1/3})/3-x^{1/2}-4

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derivative of

f (x) = 4 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 4 x + 1

2 x + \frac{16 x ^{\frac{1}{3}}}{3} - x^{\frac{1}{2}} - 4

Solution steps

\frac{d}{d x} (4 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 4 x + 1)

Simplify

4 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 4 x + 1 : 4 x^{\frac{4}{3}} - \frac{2 x ^{\frac{3}{2}}}{3} + x^{2} - 4 x + 1

= \frac{d}{d x} (4 x^{\frac{4}{3}} - \frac{2 x ^{\frac{3}{2}}}{3} + x^{2} - 4 x + 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (4 x^{\frac{4}{3}}) - \frac{d}{d x} (\frac{2 x ^{\frac{3}{2}}}{3}) + \frac{d}{d x} (x^{2}) - \frac{d}{d x} (4 x) + \frac{d}{d x} (1)

\frac{d}{d x} (4 x^{\frac{4}{3}}) = \frac{16 x ^{\frac{1}{3}}}{3}

\frac{d}{d x} (\frac{2 x ^{\frac{3}{2}}}{3}) = x^{\frac{1}{2}}

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} (4 x) = 4

\frac{d}{d x} (1) = 0

= \frac{16 x ^{\frac{1}{3}}}{3} - x^{\frac{1}{2}} + 2 x - 4 + 0

Simplify

= 2 x + \frac{16 x ^{\frac{1}{3}}}{3} - x^{\frac{1}{2}} - 4

n = 0 \sum \infty 2 \cdot 0. 2^{n}

x \to 0 lim (\frac{5 ^{x + 1} - 5}{x})

\int_{0}^{\infty} 8 sin^{2} (x) d x

\int arctan (d) x d x

\frac{d}{d θ} (\frac{2}{1 - sin ( θ )})

What is the derivative of f(x)=4x^{4/3}-2/3 x^{3/2}+x^2-4x+1 ?
The derivative of f(x)=4x^{4/3}-2/3 x^{3/2}+x^2-4x+1 is 2x+(16x^{1/3})/3-x^{1/2}-4
What is the first derivative of f(x)=4x^{4/3}-2/3 x^{3/2}+x^2-4x+1 ?
The first derivative of f(x)=4x^{4/3}-2/3 x^{3/2}+x^2-4x+1 is 2x+(16x^{1/3})/3-x^{1/2}-4